A positive charge q of mass m is moving along | vedclass

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(C) The magnetic field provides the necessary centripetal force for the particle to move in a semi-circular path of diameter $d$,so the radius $R = \f

Vedclass · Accepted Answer

$B = \frac{2mv}{qd}$ and $\Delta t = \frac{\pi d}{2v}$

Vedclass · Answer

(C) The magnetic field provides the necessary centripetal force for the particle to move in a semi-circular path of diameter $d$,so the radius $R = \frac{d}{2}$.Using the centripetal force formula: $Bqv = \frac{mv^2}{R} = \frac{mv^2}{d/2}$.Solving for $B$: $B = \frac{2mv}{qd}$.The time $\Delta t$ taken to complete a semi-circle is half the time period of a full circular orbit.The time period $T = \frac{2\pi m}{Bq}$.Thus,$\Delta t = \frac{T}{2} = \frac{\pi m}{Bq}$.Substituting $B = \frac{2mv}{qd}$ into the expression for $\Delta t$:$\Delta t = \frac{\pi m}{(2mv/qd)q} = \frac{\pi d}{2v}$.

$A$ positive charge $q$ of mass $m$ is moving along the $+x$ axis with velocity $v$. We wish to apply a uniform magnetic field $B$ for time $\Delta t$ so that the charge reverses its direction,crossing the $y$ axis at a distance $d$. Then:

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